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Keeping the Rearrangement Theorem in mind. The first part of the answer is how many ways we can write a multiplication table of size nxn that have all it rows and columns with no repeated element.

Second part to the answer is out of the number that was calculated in first part how many isomorphic groups we can eliminate?

for example we can prove that all columns permutations will yield isomorphic groups by mapping the first rows of any of those with the permutation done.

what is the general formula for n in the first part? And is there is a formula for the second part, or it all done by brute force? if so how far is known?

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    If you can classify all finite groups, or even just count them, you will be a very famous mathematician indeed. – 2017-02-10
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    It is [known](https://en.wikipedia.org/wiki/P-group#Prevalence) that the number of non-isomorphic $p$-groups of order $p^k$ is $p^{\frac{2}{27} k^3 + O(k^{8 / 3})}$. However, as far as I know there isn't even an asymptotic formula for the number of non-isomorphic general groups of a given order. – 2017-02-10
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    Highly relevant: https://oeis.org/A000001 – 2017-02-10

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See Wikipedia for the number of Latin squares (up to the natural equivalence, which Wikipedia called "reduction"). In particular, for an $n\times n$ grid this is greater than or equal to $\frac{(n!)^{2n}}{n^{n^2}}$.

For example, there are 5 groups of order $8$, but quite a few $8\times8$ Latin squares (535,281,401,856, to be precise). There is only 1 group of order $11$, but the number of $11\times11$ Latin squares has increased: 5,363,937,773,277,371,298,119,673,540,771,840.

In particular, the number of Latin squares strictly increase as $n$ increases, while we know that the number of groups jumps around a lot. Qudit's comment then makes a lot of sense - the number of groups of order $p^k$ strictly increases as $k$ increases, so start thinking about $p$-groups (in particular, $2$-groups as there is a folk conjecture that "almost all finite groups are $2$-groups").